Justify that $u(x,y)\overset{y\uparrow\infty}{\longrightarrow}0$ implies $\mathcal{F}_{x}(u)(\alpha,y)\overset{y\uparrow\infty}{\longrightarrow}0$ I'm having my Fourier-Analysis (and measure theory) exam soon, and one problem that will certainly appear is of the following form:
You have a partial differential equation (e.g. the heat equation) in say one spatial (and one time) dimension. Then you use either Fourier series, or Fourier transform to formally, meaning without caring for convergence issues etc., compute the general solution, and then you prove that the thereby found solution is $C^{\infty}$ in the interior of the domain, $C^0$ to the border, and that under some strong conditions for the initial conditions, also they are satisfied (like $u(x,0)=f(x)\forall x$ and then you suppose $f\in C^4$).
I'm mostly at ease with this procedure, but I don't like the part of formal calculation that much. I see that most of the things can be justified a posteriori with e.g. unicity of solutions, but there is one thing I cannot wrap my head arround:
Say you want to find $u:\overline{\Omega}\to\mathbb{R}$, where $\Omega=\mathbb{R}\times\mathbb{R}_{>0}$ such that 
$$\triangle u(x,y)=0\qquad\forall(x,y)\in\Omega\\ 
u(x,0)=f(x)\qquad\forall x\in\mathbb{R}\\ 
u(x,y)\overset{y\uparrow\infty}{\longrightarrow}0\qquad\forall x\in\mathbb{R}$$ where $f\in C^0(\mathbb{R})\cap L^1(\mathbb{R})$ and $|\hat{f}|\in L^1(\mathbb{R})$. Now our professor told us that the idea is to formally introduce 
$$
v(\alpha,y)=\mathcal{F}_{x}(u)(\alpha,y):=\int_{-\infty}^{+\infty}u(x,y)\cdot e^{-2\pi i\alpha x}dx
$$
and translate the conditions on $u$ to conditions on $v$. He then translates the third condition to $v(\alpha,y)\overset{y\uparrow\infty}{\longrightarrow}0$. How is this done? I see that it would follow directly if the familly of functions $\{u(\cdot,y)\}_{y>0}$ would be uniformly bounded by a $L^1(\mathbb{R})$ function (by dominated convergence). Can this somehow be deduced from the conditions? Or is this just another "formal" argument, with justification only a posteriori?
 A: One of the difficult parts of this problem is that there is not just one way to define conditions leading to uniqueness. For example, if $g(x,y)$ is harmonic in the upper half plane where it is uniformly bounded by a constant $M$, then $\lim_{y\downarrow 0} g(x,y)=f(x)$ exists for almost every $x$ and $g$ is the Poisson integral of this boundary function $f$:
$$
          g(x,y) = \frac{1}{\pi}\int_{-\infty}^{\infty}\frac{f(x')y}{(x'-x)^2+y^2}dx'.
$$
Furthermore, $g$ is the Poisson integral of only one such bounded measurable function $f$, up to an a.e. equivalence dictated by the Lebesgue integral. This is very difficult to prove on the upper half-plane; in fact, I believe that the only known proof is carried out by recasting the problem on the unit disk instead, and then transforming to the upper half-plane through a change of variables, no doubt in part because of how much more tedious the proof of uniqueness is for case of a bounded boundary function.
So, the first thing to note in your problem is that one could argue that there are extra conditions inserted in the uniqueness statement you were given. Boundedness would be enough to formulate a uniqueness statement, and vanishing limits as $y\rightarrow\infty$ is just icing on the cake that is not really needed.
Boundedness eliminates the solution $g(x,y)=y$ which vanishes at $y=0$ and is unbounded on the upper half plane. This solution is eliminated by imposing $0$ limits as $y\rightarrow\infty$, but boundedness is more than enough to eliminate this solution that causes non-uniqueness. Adding the extra assumption that $f \in L^1$ makes it easier to deal with the problem through the Fourier transform, even though nothing is added beyond the boundedness assumption, at least from a uniqueness perspective.
Boundedness is implicitly found in the assumptions you were given because $f$ is the inverse Fourier transform of $\hat{f}$. The added conditions are designed to allow you to Fourier transform the problem and come up with a unique transform.
From my perspective, it is the non-uniqueness of the uniqueness conditions that is confusing. And it really is a difficult issue when dealing with PDEs, even equations of a classical nature, and especially on unbounded regions. The conditions stated for you are those that make it convenient to be able to come up with a unique Fourier transform of the solution.
Note: The conditions you were given do not appear to be strong enough to deal with the Fourier transform directly. For example, you were not given enough information to know that $u(x,y)$ is in $L^1$ for each $y$, which makes it hard to justify that the Fourier integral exists. Typical conditions in this regard are
$$
             \sup_{y > 0}\int_{-\infty}^{\infty}|u(x,y)|dx < \infty, \\
          \lim_{y'\rightarrow y}\int_{-\infty}^{\infty}|u(x,y)-u(x,y')|dx = 0,\; y > 0, \\
    \lim_{y\downarrow 0}\int_{-\infty}^{\infty}|u(x,y)-f(x)|dx = 0.
$$
These conditions are satisfied by the Fourier solution.
